Cho bieu thuc A= 2 can cua x / can cua x (chi 1 mik x) - 2 + can cua 1 mik x +2/ 3-can cua 1 mik x + (4 nhan can cua 1 mk x - 1) / x-5 nhan can 1 mk x + 6.Rut gon bieu thuc A(giai chi tiet kem dap an, co dieu kien xac dinh)

$$\eqalign{ & A = {{2\sqrt x } \over {\sqrt x - 2}} + {{\sqrt x + 2} \over {3 - \sqrt x }} + {{4\left( {\sqrt x - 1} \right)} \over {x - 5\sqrt x + 6}}\,\,\left( {x \ge 0;\,\,x \ne 4;\,\,x \ne 9} \right) \cr & A = {{2\sqrt x } \over {\sqrt x - 2}} - {{\sqrt x + 2} \over {\sqrt x - 3}} + {{4\left( {\sqrt x - 1} \right)} \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} \cr & A = {{2\sqrt x \left( {\sqrt x - 3} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + 4\sqrt x - 4} \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} \cr & A = {{2x - 6\sqrt x - \left( {x - 4} \right) + 4\sqrt x - 4} \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} \cr & A = {{2x - 6\sqrt x - x + 4 + 4\sqrt x - 4} \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} \cr & A = {{x - 2\sqrt x } \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} \cr & A = {{\sqrt x \left( {\sqrt x - 2} \right)} \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} \cr & A = {{\sqrt x } \over {\sqrt x - 3}} \cr} $$